How do you get the sides of the triangle when 3 angles are given and its perimeter?

Answer 1

You need to use the sine rule.

If the three angles are A, B and C and the sides opposite them are named a, b and c then, by the sine rule,

a/sin(A) = b/sin(b) = c/sin(C)

Therefore

b = a*sin(B)/sin(A) = a*y where y = sin(B)/sin(A) can be calculated

and

c = a*sin(C)/sin(A) = a*z where z = sin(C)/sin(A) can be calculated.

then perimeter = p = a + b + c = a + ay + az = a*(1 + y + z)

therefore a = p/(1 + y + z)

or a = p/[1 + sin(B)/sin(A) + sin(C)/sin(A)].

Everything on the right hand side is known and so a can be calculated.

Once that has been done, b = a*y and c = a*z.